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Bavarian

One's examination of this problem is based on one of six parts involved; and it is from these options that the confusion stems, because it is easy to lose track of which part is important. The following ratios are relevant:

# Liquid T / Total

# Liquid F / Total

# Liquid T / Liquid F

# Liquid F / Liquid T

# Liquid T / 3 cups

# Liquid F / 3 cups

At no point would the mixtures have a 50-50 distribution for the same reason that a drop of red dye in a glass of water is not distributed 50-50. Problems such as these do assume an "uniform distribution" but this refers to the idea that a spoonful of the mixture from the top, bottom, or middle of the mixture would all have the same ratio of Liquid T / Liquid F.

At first, you pour 3 cl from glass A in glass B, and after mixing you pour 3 cl from glass B back into A.

Ergo, in the end, both glasses still contain 10 cl, just like at the start of the experiment. Therefor the amount of fernet in the glass of tonic must be exactly the same as the amount of tonic in the glass of fernet.

One might be inclined to look for a complicated calculation, although it's as simple as stated above

BoilingOil

Why "must" it be exactly the same as the amount of tonic in the glass of fernet. What proof do you have? The mathematics will validate what you have only claimed to be true.

Why "must" it be exactly the same as the amount of tonic in the glass of fernet. What proof do you have? The mathematics will validate what you have only claimed to be true.

Don't over think this one. It was eloquently explained earlier:

no math needed. the same amount of liquid is in each glass. so whatever tonic that isnt in the tonic glass is in the other glass, and vice versaassuming nothing spilled

It's a closed system. There's a total of 20cl of liquid, 10 of each type, so no matter how it's mixed, whatever isn't in one must be in the other. Since per instructions, after pouring there is again 10 cl in each container, whatever tonic is now in the fernet flask must be equal to the amount of fernet that is now in the tonic flask.

Isn't that what we do here? While the proof offered by unreality, yourself, and others is not incorrect, it is not rigorous enough. An extremely beautiful proof could be written using your idea, a better proof than the one I was imagining, but unless your thinking is expressed in a more rigorous language than the common sense approach you've adopted, it will not possess the beauty it deserves.

It doesn't matter what the amount is you pour as long as each pour is the same. Keep in mind this doesn't make the glasses contain the same solution it only give equal parts of the opposite liquid in each one.

being a chemist when you say how much I think concentration, Assuming even concentrations, the math does work out
you wind up with 0.2307692 the original concentration of tonic in the fernet with the fernet maintaining 0.7692307 times its original concentration. the exact reverse is true for the tonic. if you want to check it for yourself the only equation you need is: concentration 1 * volume 1 = concentration 2 * volume 2 where concentration 1 is your initial concentration and volume 1 is the volume of the amount you pour out then concentration 2 is your unknown and volume 2 is the total volume in the final glass. give it a try.

First to anyone applying the C1V1 = C2V2 formula, it doesn't work - you're not only changing V2, but also C2 (i.e. I just converted everything to Molarity because I like working with moles - 10 moles/1L = x moles/1.3L is the natural route to go, BUT if you solve for unknown concentration "X", you're neglecting the amount of tonic poured into the fernet).

It's been too long since General Chemistry for me (ok, so maybe the truth is I've just killed too many brain cells), so I can't work this out using solubility rules. HOWEVER, using logic you pour tonic into fernet. You then pour the same volume of fernet solution back into the tonic, but this solution also contains tonic. You are therefore removing some of the original tonic poured into the fernet (while at same time pouring same amount of fernet into tonic as tonic originally poured into fernet).